THE previous chapter contained derivations of the relationships for the conservation

Transcription

1 Chapter 6 SOLUTION OF VISCOUS-FLOW PROBLEMS 6.1 Introduction THE previous chapter contained derivations of the relationships for the conservation of mass and momentum the equations of motion in rectangular, clindrical, and spherical coordinates. All the experimental evidence indicates that these are indeed the most fundamental equations of fluid mechanics, and that in principle the govern an situation involving the flow of a Newtonian fluid. Unfortunatel, because of their all-embracing qualit, their solution in analtical terms is difficult or impossible except for relativel simple situations. However, it is important to be aware of these Navier-Stokes equations, for the following reasons: 1. The lead to the analtical and exact solution of some simple, et important problems, as will be demonstrated b examples in this chapter. 2. The form the basis for further work in other areas of chemical engineering. 3. If a few realistic simplifing assumptions are made, the can often lead to approximate solutions that are eminentl acceptable for man engineering purposes. Representative examples occur in the stud of boundar laers, waves, lubrication, coating of substrates with films, and inviscid (irrotational) flow. 4. With the aid of more sophisticated techniques, such as those involving power series and asmptotic expansions, and particularl computer-implemented numerical methods, the can lead to the solution of moderatel or highl advanced problems, such as those involving injection-molding of polmers and even the incredibl difficult problem of weather prediction. The following sections present exact solutions of the equations of motion for several relativel simple problems in rectangular, clindrical, and spherical coordinates. Throughout, unless otherwise stated, the flow is assumed to be stead, laminar and Newtonian, with constant densit and viscosit. Although these assumptions are necessar in order to obtain solutions, the are nevertheless realistic in man cases. All of the examples in this chapter are characterized b low Renolds numbers. That is, the viscous forces are much more important than the inertial forces, and 272

2 6.1 Introduction 273 are usuall counterbalanced b pressure or gravitational effects. Tpical applications occur at low flow rates and in the flow of high-viscosit polmers. Situations in which viscous effects are relativel unimportant will be discussed in Chapter 7. Solution procedure. The general procedure for solving each problem involves the following steps: 1. Make reasonable simplifing assumptions. Almost all of the cases treated here will involve stead incompressible flowofanewtonian fluid in a single coordinate direction. Further, gravit ma or ma not be important, and a certain amount of smmetr ma be apparent. 2. Write down the equations of motion both mass (continuit) and momentum balances and simplif them according to the assumptions made previousl, striking out terms that are zero. Tpicall, onl a ver few terms perhaps onl one in some cases will remain in each differential equation. The simplified continuit equation usuall ields information that can subsequentl be used to simplif the momentum equations. 3. Integrate the simplified equations in order to obtain expressions for the dependent variables such as velocities and pressure. These expressions will usuall contain some, as et, arbitrar constants tpicall two for the velocities (since the appear in second-order derivatives in the momentum equations) and one for the pressure (since it appears onl in a first-order derivative). 4. Invoke the boundar conditions in order to evaluate the constants appearing in the previous step. For pressure, such a condition usuall amounts to a specified pressure at a certain location at the inlet of a pipe, or at a free surface exposed to the atmosphere, for example. For the velocities, these conditions fall into either of the following classifications: (a) Continuit of the velocit, amounting to a no-slip condition. Thus, the velocit of the fluid in contact with a solid surface tpicall equals the velocit of that surface zero if the surface is stationar. 1 And, for the few cases in which one fluid (A, sa) is in contact with another immiscible fluid (B), the velocit in fluid A equals the velocit in fluid B at the common interface. (b) Continuit of the shear stress, usuall between two fluids A and B, leading to the product of viscosit and a velocit gradient having the same value at the common interface, whether in fluid A or B. If fluid A is a liquid, and fluid B is a relativel stagnant gas, which because of its low viscosit is incapable of sustaining an significant shear stress, then the common shear stress is effectivel zero. 5. At this stage, the problem is essentiall solved for the pressure and velocities. Finall, if desired, shear-stress distributions can be derived b differentiating 1 In a few exceptional situations there ma be lack of adhesion between the fluid and surface, in which case slip can occur.

3 274 Chapter 6 Solution of Viscous-Flow Problems the velocities in order to obtain the velocit gradients; numerical predictions of process variables can also be made. Tpesofflow. Two broad classes of viscous flow will be illustrated in this chapter: 1. Poiseuille flow, in which an applied pressure difference causes fluid motion between stationar surfaces. 2. Couette flow, in which a moving surface drags adjacent fluid along with it and thereb imparts a motion to the rest of the fluid. Occasionall, it is possible to have both tpes of motion occurring simultaneousl, as in the screw extruder analzed in Example Solution of the Equations of Motion in Rectangular Coordinates The remainder of this chapter consists almost entirel of a series of worked examples, illustrating the above steps for solving viscous-flow problems. Example 6.1 Flow Between Parallel Plates Fig. E6.1.1 shows the flow of a fluid of viscosit µ, which flows in the x direction between two rectangular plates, whose width is ver large in the z direction when compared to their separation in the direction. Such a situation could occur in a die when a polmer is being extruded at the exit into a sheet, which is subsequentl cooled and solidified. Determine the relationship between the flow rate and the pressure drop between the inlet and exit, together with several other quantities of interest. Top plate Large width Inlet Coordinate sstem x z Fluid Narrow gap, 2d Bottom plate Fig. E6.1.1 Geometr for flow through a rectangular duct. The spacing between the plates is exaggerated in relation to their length. Simplifing assumptions. The situation is analzed b referring to a cross section of the duct, shown in Fig. E6.1.2, taken at an fixed value of z. Let the depth be 2d (±d above and below the centerline or axis of smmetr = 0), and the length L. Note that the motion is of the Poiseuille tpe, since it is caused b the applied pressure difference (p 1 p 2 ). Make the following realistic assumptions about the flow: Exit

4 6.2 Solution of the Equations of Motion in Rectangular Coordinates As alread stated, it is stead and Newtonian, with constant densit and viscosit. (These assumptions will often be taken for granted, and not restated, in later problems.) 2. There is onl one nonzero velocit component that in the direction of flow, v x. Thus, v = v z =0. 3. Since, in comparison with their spacing, 2d, the plates extend for a ver long distance in the z direction, all locations in this direction appear essentiall identical to one another. In particular, there is no variation of the velocit in the z direction, so that v x / z =0. 4. Gravit acts verticall downwards; hence, g = g and g x = g z =0. 5. The velocit is zero in contact with the plates, so that v x =0at = ±d. Continuit. Start b examining the general continuit equation, (5.48): ρ t + (ρv x) + (ρv ) + (ρv z) =0, (5.48) x z which, in view of the constant-densit assumption, simplifies to Eqn. (5.52): v x x + v + v z z =0. (5.52) But since v = v z =0: v x x =0, (E6.1.1) so v x is independent of the distance from the inlet, and the velocit profile will appear the same for all values of x. Since v x / z = 0 (assumption 3), it follows that v x = v x () is a function of onl. Inlet Wall Velocit profile Exit d v x p 1 x p 2 Axis of smmetr d Wall Fig. E6.1.2 L Geometr for flow through a rectangular duct.

5 276 Chapter 6 Solution of Viscous-Flow Problems Momentum balances. With the stated assumptions of a Newtonian fluid with constant densit and viscosit, Eqn. (5.73) gives the x,, and z momentum balances: ( ) vx ρ t + v v x x x + v v x + v v x z = p ( ) 2 z x + µ v x x + 2 v x v x + ρg 2 z 2 x, ( ) v ρ t + v v x x + v v + v v z = p ( ) 2 z + µ v x + 2 v v + ρg 2 z 2, ( vz ρ t + v v z x x + v v z + v z v z z ) = p z + µ ( 2 v z x v z v z z 2 ) + ρg z. With v = v z = 0 (from assumption 2), v x / x = 0 [from the simplified continuit equation, (E6.1.1)], g = g, g x = g z = 0 (assumption 4), and stead flow (assumption 1), these momentum balances simplif enormousl, to: µ 2 v x 2 p = p x, = ρg, p z =0. (E6.1.2) (E6.1.3) (E6.1.4) Pressure distribution. The last of the simplified momentum balances, Eqn. (E6.1.4), indicates no variation of the pressure across the width of the sstem (in the z direction), which is hardl a surprising result. When integrated, the second simplified momentum balance, Eqn. (E6.1.3), predicts that the pressure varies according to: p = ρg d + f(x) = ρg + f(x). (E6.1.5) Observe carefull that since a partial differential equation is being integrated, we obtain not a constant of integration, but a function of integration, f(x). Assume to be verified later that p/ x is constant, so that the centerline pressure (at = 0) is given b a linear function of the form: p =0 = a + bx. (E6.1.6) The constants a and b ma be determined from the inlet and exit centerline pressures: x =0: p = p 1 = a, (E6.1.7) x = L : p = p 2 = a + bl, (E6.1.8)

6 6.2 Solution of the Equations of Motion in Rectangular Coordinates 277 leading to: a = p 1, b = p 1 p 2 L. (E6.1.9) Thus, the centerline pressure declines linearl from p 1 at the inlet to p 2 at the exit: f(x) =p 1 x L (p 1 p 2 ), (E6.1.10) so that the complete pressure distribution is p = p 1 x L (p 1 p 2 ) ρg. (E6.1.11) That is, the pressure declines linearl, both from the bottom plate to the top plate, and also from the inlet to the exit. In the majorit of applications, 2d L, and the relativel small pressure variation in the direction is usuall ignored. Thus, p 1 and p 2, although strictl the centerline values, are tpicall referred to as the inlet and exit pressures, respectivel. Velocit profile. Since, from Eqn. (E6.1.1), v x does not depend on x, 2 v x / 2 appearing in Eqn. (E6.1.2) becomes a total derivative, so this equation can be rewritten as: µ d2 v x d 2 = p x, (E6.1.12) which is a second-order ordinar differential equation, in which the pressure gradient will be shown to be uniform between the inlet and exit, being given b: p x = p 1 p 2 L. (E6.1.13) [A minus sign is used on the left-hand side, since p/ x is negative, thus rendering both sides of Eqn. (E6.1.13) as positive quantities.] Equation (E6.1.12) can be integrated twice, in turn, to ield an expression for the velocit. After multiplication through b d, a first integration gives: ( ) ( ) d 2 v x d d d = dvx 1 p d = d, 2 d d µ x dv x d = 1 ( ) p + c 1. (E6.1.14) µ x A second integration, of Eqn. (E6.1.14), ields: [ ( ) ] dvx 1 p d d = + c 1 d, µ x v x = 1 ( ) p 2 + c 1 + c 2. 2µ x (E6.1.15)

7 278 Chapter 6 Solution of Viscous-Flow Problems The two constants of integration, c 1 and c 2, are determined b invoking the boundar conditions: leading to: =0: dv x d =0, (E6.1.16) = d : v x =0, (E6.1.17) c 1 =0, c 2 = 1 2µ ( ) p d 2. x (E6.1.18) Eqns. (E6.1.15) and (E6.1.18) then furnish the velocit profile: v x = 1 ( p ) (d 2 2 ), (E6.1.19) 2µ x in which p/ x and (d 2 2 ) are both positive quantities. The velocit profile is parabolic in shape, and is shown in Fig. E Alternative integration procedure. Observe that we have used indefinite integrals in the above solution, and have emploed the boundar conditions to determine the constants of integration. An alternative approach would again be to integrate Eqn. (E6.1.12) twice, but now to involve definite integrals b inserting the boundar conditions as limits of integration. Thus, b separating variables, integrating once, and noting from smmetr about the centerline that dv x /d =0at = 0, we obtain: or: dvx/d ( ) dvx µ d = p d, 0 d x 0 dv x d = 1 µ (E6.1.20) ( ) p. (E6.1.21) x A second integration, noting that v x =0at = d (zero velocit in contact with the upper plate the no-slip condition) ields: vx 0 dv x = 1 µ ( ) p d. x d (E6.1.22) That is: v x = 1 ( p ) (d 2 2 ), (E6.1.23) 2µ x in which two minus signs have been introduced into the right-hand side in order to make quantities in both parentheses positive. This result is identical to the earlier

8 6.2 Solution of the Equations of Motion in Rectangular Coordinates 279 Eqn. (E6.1.19). The student is urged to become familiar with both procedures, before deciding on the one that is individuall best suited. Also, the reader who is troubled b the assumption of smmetr of v x about the centerline (and b never using the fact that v x =0at = d), should be reassured b an alternative approach, starting from Eqn. (E6.1.15): 0 d v x = 1 2µ d ( p x ) 2 + c 1 + c 2. (E6.1.24) Application of the two boundar conditions, v x =0at = ±d, gives c 1 =0, c 2 = 1 ( ) p d 2, (E6.1.25) 2µ x leading again to the velocit profile of Eqn. (E6.1.19) without the assumption of smmetr. Volumetric flow rate. Integration of the velocit profile ields an expression for the volumetric flow rate Q per unit width of the sstem. Observe first that the differential flow rate through an element of depth d is dq = v x d, so that: Q d d ( 1 Q = dq = v x d = p ) ( (d 2 2 ) d = 2d3 p ). (E6.1.26) 2µ x 3µ x Since from an overall macroscopic balance Q is constant, it follows that p/ x is also constant, independent of distance x; the assumptions made in Eqns. (E6.1.6) and (E6.1.13) are therefore verified. The mean velocit is the total flow rate per unit depth: v xm = Q 2d = d2 3µ ( p ), (E6.1.27) x and is therefore two-thirds of the maximum velocit, v xmax, which occurs at the centerline, =0. Pressure at inlet Wall Shear-stress distribution Pressure at exit Inlet Exit Axis of smmetr p 1 x p 2 Wall Fig. E6.1.3 Pressure and shear-stress distributions.

9 280 Chapter 6 Solution of Viscous-Flow Problems Shear-stress distribution. Finall, the shear-stress distribution is obtained b emploing Eqn. (5.60): ( vx τ x = µ + v ). (5.60) x B substituting for v x from Eqn. (E6.1.15) and recognizing that v = 0, the shear stress is: ( τ x = p ). (E6.1.28) x Referring back to the sign convention expressed in Fig. 5.13, the first minus sign in Eqn. (E6.1.28) indicates for positive that the fluid in the region of greater is acting on the region of lesser in the negative x direction, thus tring to retard the fluid between it and the centerline, and acting against the pressure gradient. Representative distributions of pressure and shear stress, from Eqns. (E6.1.11) and (E6.1.28), are sketched in Fig. E More precisel, the arrows at the left and right show the external pressure forces acting on the fluid contained between x = 0 and x = L. 6.3 Alternative Solution Using a Shell Balance Because the flow between parallel plates was the first problem to be examined, the analsis in Example 6.1 was purposel ver thorough, extracting the last ounce of information. In man other applications, the velocit profile and the flow rate ma be the onl quantities of prime importance. On the average, therefore, subsequent examples in this chapter will be shorter, concentrating on certain features and ignoring others. The problem of Example 6.1 was solved b starting with the completel general equations of motion and then simplifing them. An alternative approach involves a direct momentum balance on a differential element of fluid a shell as illustrated in Example 6.2. Example 6.2 Shell Balance for Flow Between Parallel Plates Emplo the shell-balance approach to solve the same problem that was studied in Example 6.1. Assumptions. The necessar shell is in realit a differential element of fluid, as shown in Fig. E6.2. The element, which has dimensions of dx and d in the plane of the diagram, extends for a depth of dz (an other length ma be taken) normal to the plane of the diagram. If, for the present, the element is taken to be a sstem that is fixed in space, there are three different tpes of rate of x-momentum transfer to it:

10 6.3 Alternative Solution Using a Shell Balance A convective transfer of ρv 2 x d dz in through the left-hand face, and an identical amount out through the right-hand face. Note here that we have implicitl assumed the consequences of the continuit equation, expressed in Eqn. (E6.1.1), that v x is constant along the duct. 2. Pressure forces on the left- and right-hand faces. The latter will be smaller, because p/ x is negative in realit. 3. Shear stresses on the lower and upper faces. Observe that the directions of the arrows conform strictl to the sign convention established in Section 5.6. t x + t x d r v x 2 p d dx rv x 2 p + p x dx x t x Fig. E6.2 Momentum balance on a fluid element. A momentum balance on the element, which is not accelerating, gives: ( ρv 2 x d dz ρv 2 x d dz + pddz p + p ) } {{ } } {{ } } {{ } x dx d dz In Out Left } {{ } Net convective transfer ( + } {{ } Right } {{ } Net pressure force ) τ x + τ x d dx dz } {{ } Upper τ x dx dz } {{ } Lower =0. (E6.2.1) } {{ } Net shear force The usual cancellations can be made, resulting in: dτ x d = p x, (E6.2.2) in which the total derivative recognizes that the shear stress depends onl on and not on x. Substitution for τ x from Eqn. (5.60) with v = 0 gives: µ d2 v x d 2 = p x, (E6.2.3) which is identical with Eqn. (E6.1.12) that was derived from the simplified Navier- Stokes equations. The remainder of the development then proceeds as in the previous example. Note that the convective terms can be sidestepped entirel if the momentum balance is performed on an element that is chosen to be moving with the fluid, in which case there is no flow either into or out of it.

11 282 Chapter 6 Solution of Viscous-Flow Problems The choice of approach simplifing the full equations of motion, or performing a shell balance is ver much a personal one, and we have generall opted for the former. The application of the Navier-Stokes equations, which are admittedl rather complicated, has the advantages of not reinventing the (momentum balance) wheel for each problem, and also of assuring us that no terms are omitted. Conversel, a shell balance has the merits of relative simplicit, although it ma be quite difficult to perform convincingl for an element with curved sides, as would occur for the problem in spherical coordinates discussed in Example 6.6. This section concludes with another example problem, which illustrates the application of two further boundar conditions for a liquid, one involving it in contact with a moving surface, and the other at a gas/liquid interface where there is a condition of zero shear. Example 6.3 Film Flow on a Moving Substrate Fig. E6.3.1 shows a coating experiment involving a flat photographic film that is being pulled up from a processing bath b rollers with a stead velocit U at an angle θ to the horizontal. As the film leaves the bath, it entrains some liquid, and in this particular experiment it has reached the stage where: (a) the velocit of the liquid in contact with the film is v x = U at = 0, (b) the thickness of the liquid is constant at a value δ, and (c) there is no net flow of liquid (as much is being pulled up b the film as is falling back b gravit). (Clearl, if the film were to retain a permanent coating, a net upwards flow of liquid would be needed.) q x Air Liquid coating Moving photographic film Fig. E6.3.1 Liquid coating on a photographic film. Perform the following tasks: 1. Write down the differential mass balance and simplif it. 2. Write down the differential momentum balances in the x and directions. What are the values of g x and g in terms of g and θ? Simplif the momentum balances as much as possible. 3. From the simplified momentum balance, derive an expression for the pressure p as a function of, ρ, δ, g, and θ, and hence demonstrate that p/ x =0. Assume that the pressure in the surrounding air is zero everwhere. 4. From the simplified x momentum balance, assuming that the air exerts a negligible shear stress τ x on the surface of the liquid at = δ, derive an d U g

12 6.3 Alternative Solution Using a Shell Balance 283 expression for the liquid velocit v x as a function of U,, δ, and α, where α = ρg sin θ/µ. 5. Also derive an expression for the net liquid flow rate Q (per unit width, normal to the plane of Fig. E6.3.1) in terms of U, δ, and α. Noting that Q = 0, obtain an expression for the film thickness δ in terms of U and α. 6. Sketch the velocit profile v x, labeling all important features. Assumptions and continuit. The following assumptions are reasonable: 1. The flow is stead and Newtonian, with constant densit ρ and viscosit µ. 2. The z direction, normal to the plane of the diagram, ma be disregarded entirel. Thus, not onl is v z zero, but all derivatives with respect to z, such as v x / z, are also zero. 3. There is onl one nonzero velocit component, namel, that in the direction of motion of the photographic film, v x. Thus, v = v z =0. 4. Gravit acts verticall downwards. Because of the constant-densit assumption, the continuit equation, (5.48), simplifies, as before, to: v x x + v + v z z =0. (E6.3.1) But since v = v z = 0, it follows that v x x =0. (E6.3.2) so v x is independent of distance x along the film. Further, v x does not depend on z (assumption 2); thus, the velocit profile v x = v x () depends onl on and will appear the same for all values of x. Momentum balances. With the stated assumptions of a Newtonian fluid with constant densit and viscosit, Eqn. (5.73) gives the x and momentum balances: ( vx ρ t + v v x x x + v v x + v z ( v ρ t + v v x x + v v + v z ) v x = p z ) v z ( 2 x + µ v x x 2 = p + µ ( 2 v x v x v 2 ) + 2 v x + ρg z 2 x, + 2 v z 2 ) + ρg, Noting that g x = g sin θ and g = g cos θ, these momentum balances simplif to: p x + ρg sin θ = v x µ 2, 2 p = ρg cos θ. (E6.3.4) (E6.3.3)

13 284 Chapter 6 Solution of Viscous-Flow Problems Integration of Eqn. (E6.3.4), between the free surface at = δ (where the gauge pressure is zero) and an arbitrar location (where the pressure is p) gives: p 0 dp = ρg cos θ δ d + f(x), (E6.3.5) so that: p = ρg cos θ(δ )+f(x). (E6.3.6) Note that since a partial differential equation is being integrated, a function of integration, f(x), is again introduced. Another wa of looking at it is to observe that if Eqn. (E6.3.6) is differentiated with respect to, we would recover the original equation, (E6.3.4), because f(x)/ = 0. However, since p =0at = δ (the air/liquid interface) for all values of x, the function f(x) must be zero. Hence, the pressure distribution: p =(δ )ρg cos θ, (E6.3.7) shows that p is not a function of x. In view of this last result, we ma now substitute p/ x = 0 into the x- momentum balance, Eqn. (E6.3.3), which becomes: d 2 v x d 2 = ρg µ sin θ = α, (E6.3.8) in which the constant α has been introduced to denote ρg sin θ/µ. Observe that the second derivative of the velocit now appears as a total derivative, since v x depends on onl. A first integration of Eqn. (E6.3.8) with respect to gives: dv x d = α + c 1. (E6.3.9) The boundar condition of zero shear stress at the free surface is now invoked: ( v τ x = µ x + v ) x = µ dv x d =0. (E6.3.10) Thus, from Eqns. (E6.3.9) and (E6.3.10) at = δ, the first constant of integration can be determined: dv x d = αδ + c 1 =0, or c 1 = αδ. (E6.3.11) A second integration, of Eqn. (E6.3.9) with respect to, gives: ( ) 2 v x = α 2 δ + c 2. (E6.3.12)

14 6.4 Poiseuille and Couette Flows in Polmer Processing 285 The second constant of integration, c 2, can be determined b using the boundar condition that the liquid velocit at = 0 equals that of the moving photographic film. That is, v x = U at = 0, ielding c 2 = U; thus, the final velocit profile is: v x = U α ( δ 2 ). (E6.3.13) Observe that the velocit profile, which is parabolic, consists of two parts: 1. A constant and positive part, arising from the film velocit, U. 2. A variable and negative part, which reduces v x at increasing distances from the film and eventuall causes it to become negative. Reverse flow (downwards) Free surface: x q t x =0= dv x d Moving film: Forward flow (upwards) v x = U d Fig. E6.3.2 Velocit profile in thin liquid laer on moving photographic film for the case of zero net liquid flow rate. Exactl how much of the liquid is flowing upwards, and how much downwards, depends on the values of the variables U, δ, and α. However, we are asked to investigate the situation in which there is no net flow of liquid that is, as much is being pulled up b the film as is falling back b gravit. In this case: Q = δ 0 v x d = δ giving the thickness of the liquid film as: 0 [ ( U α δ )] d = Uδ αδ3 =0, (E6.3.14) δ = 3U α. (E6.3.15) The velocit profile for this case of Q = 0 is shown in Fig. E Poiseuille and Couette Flows in Polmer Processing The stud of polmer processing falls into the realm of the chemical engineer. First, the polmer, such as nlon, polstrene, or polethlene, is produced b

15 286 Chapter 6 Solution of Viscous-Flow Problems a chemical reaction either as a liquid or solid. (In the latter event, it would subsequentl have to be melted in order to be processed further.) Second, the polmer must be formed b suitable equipment into the desired final shape, such as a film, fiber, bottle, or other molded object. The procedures listed in Table 6.1 are tpical of those occurring in polmer processing. Table 6.1 Operation Extrusion and die flow Drawing or Spinning Injection molding Blow molding Calendering Coating Tpical Polmer-Processing Operations Description The polmer is forced, either b an applied pressure, or pump, or a rotating screw, through a narrow opening called a die in order to form a continuous sheet, filament, or tube. The polmer flows out through a narrow opening, either as a sheet or a thread, and is pulled b a roller in order to make a thinner sheet or thread. The polmer is forced under high pressure into a mold, in order to form a variet of objects, such as telephones and automobile bumpers. A balloon of polmer is expanded b the pressure of a gas in order to fill a mold. Bottles are tpicall formed b blow molding. The polmer is forced through two rotating rollers in order to form a relativel thin sheet. The nature of the surfaces of the rollers will strongl influence the final appearance of the sheet, which ma be smooth, rough, or have a pattern embossed on it. The polmer is applied as a thin film b a blade or rollers on to a substrate, such as paper or a sheet of another polmer. Since polmers are generall highl viscous, their flows can be obtained b solving the equations of motion. In this chapter, we cover the rudiments of extrusion, die flow, and drawing or spinning. The analsis of calendering and coating is considerabl more complicated, but can be rendered tractable if reasonable simplifications, known collectivel as the lubrication approximation, are made, as discussed in Chapter 8. Example 6.4 The Screw Extruder Because polmers are generall highl viscous, the often need ver high pressures to push them through dies. One such pump for achieving this is the screw

16 6.4 Poiseuille and Couette Flows in Polmer Processing 287 extruder, shown in Fig. E The polmer tpicall enters the feed hopper as pellets, and is pushed forward b the screw, which rotates at an angular velocit ω, clockwise as seen b an observer looking along the axis from the inlet to the exit. The heated barrel melts the pellets, which then become fluid as the metering section of length L 0 is encountered (where the screw radius is r and the gap between the screw and barrel is h, with h r). The screw increases the pressure of the polmer melt, which ultimatel passes to a die at the exit of the extruder. The preliminar analsis given here neglects an heat-transfer effects in the metering section, and also assumes that the polmer has a constant Newtonian viscosit µ. Feed hopper Metering section L 0 Barrel Axis of rotation w Primar heating region Screw r W q h Exit to die Flights Fig. E6.4.1 Screw extruder. The investigation is facilitated b taking the viewpoint of a hpothetical observer located on the screw, in which case the screw surface and the flights appear to be stationar, with the barrel moving with velocit V = rω at an angle θ to the flight axis, as shown in Fig. E The alternative viewpoint of an observer on the inside surface of the barrel is not ver fruitful, because not onl are the flights seen as moving boundaries, but the observations would be periodicall blocked as the flights passed over the observer! Solution Motion in two principal directions is considered: 1. Flow parallel to the flight axis, caused b a barrel velocit of V = V cos θ = rωcos θ relative to the (now effectivel stationar) flights and screw. 2. Flow normal to the flight axis, caused b a barrel velocit of V x = V sin θ = rωsin θ relative to the (stationar) flights and screw. In each case, the flow is considered one-dimensional, with end-effects caused b the presence of the flights being unimportant. A glance at Fig. E6.4.3(b) will give the general idea. Although the flow in the x-direction must reverse itself as it nears the flights, it is reasonable to assume for h W that there is a substantial central region in which the flow is essentiall in the positive or negative x-direction.

17 288 Chapter 6 Solution of Viscous-Flow Problems 1. Motion parallel to the flight axis. The reader ma wish to investigate the additional simplifing assumptions that give the -momentum balance as: p = v µd2 dz. 2 Integration twice ields the velocit profile as: v = 1 2µ ( ) p (E6.4.1) z 2 + c 1 z + c 2 = 1 ( p ) (hz z 2 ) + z rω cos θ. (E6.4.2) 2µ h } {{ } } {{ } Poiseuille flow Couette flow Here, the integration constants c 1 and c 2 have been determined in the usual wa b appling the boundar conditions: z =0: v =0; z = h : v = V = V cos θ = rω cos θ. (E6.4.3) W Flight axis V Barrel h q V Flight z V x Flight Screw Fig. E6.4.2 x Diagonal motion of barrel relative to flights. Note that the negative of the pressure gradient is given in terms of the inlet pressure p 1, the exit pressure p 2, and the total length L (= L 0 / sin θ) measured along the screw flight axis b: p = p 2 p 1 L = p 1 p 2 L, (E6.4.4) and is a negative quantit since the screw action builds up pressure and p 2 >p 1. Thus, Eqn. (E6.4.2) predicts a Poiseuille-tpe backflow (caused b the adverse pressure gradient) and a Couette-tpe forward flow (caused b the relative motion of the barrel to the screw). The combination is shown in Fig. E6.4.3(a).

18 6.4 Poiseuille and Couette Flows in Polmer Processing 289 Barrel V z Flight axis h Screw (a) Cross section along flight axis, showing velocit profile. Barrel V x z Flight Flight x Screw W (b) Cross section normal to flight axis, showing streamlines. Fig. E6.4.3 Fluid motion (a) along and (b) normal to the flight axis, as seen b an observer on the screw. The total flow rate Q of polmer melt in the direction of the flight axis is obtained b integrating the velocit between the screw and barrel, and recognizing that the width between flights is W : h ( ) Q = W v dz = Wh3 p1 p µ L 2 Whrωcos θ. (E6.4.5) } {{ } } {{ } Poiseuille Couette The actual value of Q will depend on the resistance of the die located at the extruder exit. In a hpothetical case, in which the die offers no resistance, there would be no pressure increase in the extruder (p 2 = p 1 ), leaving onl the Couette term in Eqn. (E6.4.5). For the practical situation in which the die offers significant resistance, the Poiseuille term would serve to diminish the flow rate given b the Couette term. 2. Motion normal to the flight axis. B a development ver similar to that for flow parallel to the flight axis, we obtain: p x = v x µd2 dz, (E6.4.6) 2 v x = 1 ( p ) (hz z 2 ) z rω sin θ, (E6.4.7) 2µ x h } {{ } } {{ } Poiseuille flow Couette flow h ( Q x = v x dz = h3 p ) 1 hrω sin θ =0. (E6.4.8) 0 12µ x 2 } {{ } } {{ } Poiseuille Couette

19 290 Chapter 6 Solution of Viscous-Flow Problems Here, Q x is the flow rate in the x-direction, per unit depth along the flight axis, and must equal zero, because the flights at either end of the path act as barriers. The negative of the pressure gradient is therefore: p 6µrω sin θ =, (E6.4.9) x h 2 so that the velocit profile is given b: v x = z ( h rω 2 3 z ) sin θ. h (E6.4.10) Note from Eqn. (E6.4.10) that v x is zero when either z = 0 (on the screw surface) or z/h =2/3. The reader ma wish to sketch the general appearance of v z (z). 6.5 Solution of the Equations of Motion in Clindrical Coordinates Several chemical engineering operations exhibit smmetr about an axis z and involve one or more surfaces that can be described b having a constant radius for a given value of z. Examples are flow in pipes, extrusion of fibers, and viscometers that involve flow between concentric clinders, one of which is rotating. Such cases lend themselves naturall to solution in clindrical coordinates, and two examples will now be given. Example 6.5 Flow Through an Annular Die Following the discussion of polmer processing in the previous section, now consider flow through a die that could be located at the exit of the screw extruder of Example 6.4. Consider a die that forms a tube of polmer (other shapes being sheets and filaments). In the die of length D shown in Fig. E6.5, a pressure difference p 2 p 3 causes a liquid of viscosit µ to flow steadil from left to right in the annular area between two fixed concentric clinders. Note that p 2 is chosen for the inlet pressure in order to correspond to the extruder exit pressure from Example 6.4. The inner clinder is solid, whereas the outer one is hollow; their radii are r 1 and r 2, respectivel. The problem, which could occur in the extrusion of plastic tubes, is to find the velocit profile in the annular space and the total volumetric flow rate Q. Note that clindrical coordinates are now involved. Assumptions and continuit equation. The following assumptions are realistic: 1. There is onl one nonzero velocit component, namel that in the direction of flow, v z. Thus, v r = v θ =0. 2. Gravit acts verticall downwards, so that g z =0. 3. The axial velocit is independent of the angular location; that is, v z / θ =0.

20 6.5 Solution of the Equations of Motion in Clindrical Coordinates 291 To analze the situation, again start from the continuit equation, (5.49): ρ t + 1 (ρrv r ) + 1 (ρv θ ) + (ρv z) r r r θ z which, for constant densit and v r = v θ = 0, reduces to: =0, (5.49) v z z =0, (E6.5.1) verifing that v z is independent of distance from the inlet, and that the velocit profile v z = v z (r) appears the same for all values of z. Inlet Velocit p 2 Outer wall profile p 3 r z r 1 Fixed inner clinder Fluid r 2 v z v z Exit Axis of smmetr Fig. E6.5 D Geometr for flow through an annular die. Momentum balances. There are again three momentum balances, one for each of the r, θ, and z directions. If explored, the first two of these would ultimatel lead to the pressure variation with r and θ at an cross section, which is of little interest in this problem. Therefore, we extract from Eqn. (5.75) onl the z momentum balance: ( vz ρ t + v v z r r + v ) θ v z r θ + v v z z z = p [ ( 1 z + µ r v ) z + 1 ] 2 v z r r r r 2 θ + 2 v z + ρg 2 z 2 z. (E6.5.2) With v r = v θ = 0 (from assumption 1), v z / z = 0 [from Eqn. (E6.5.1)], v z / θ = 0 (assumption 3), and g z = 0 (assumption 2), this momentum balance simplifies to: [ 1 µ r d dr ( r dv )] z = p dr z, (E6.5.3)

21 292 Chapter 6 Solution of Viscous-Flow Problems in which total derivatives are used because v z depends onl on r. Shortl, we shall prove that the pressure gradient is uniform between the die inlet and exit, being given b: p z = p 2 p 3 D, (E6.5.4) in which both sides of the equation are positive quantities. Two successive integrations of Eqn. (E6.5.3) ma then be performed, ielding: v z = 1 ( p ) r 2 + c 1 ln r + c 2. (E6.5.5) 4µ z The two constants ma be evaluated b appling the boundar conditions of zero velocit at the inner and outer walls, r = r 1 : v z =0, r = r 2 : v z =0, (E6.5.6) giving: c 1 = 1 4µ ( p ) r 2 2 r1 2 z ln(r 2 /r 1 ), c 2 = 1 ( p ) r 2 2 c 1 ln r 2. 4µ z (E6.5.7) Substitution of these values for the constants of integration into Eqn. (E6.5.5) ields the final expression for the velocit profile: v z = 1 ( p )[ ] ln(r/r1 ) 4µ z ln(r 2 /r 1 ) (r2 2 r1) 2 (r 2 r1) 2, (E6.5.8) which is sketched in Fig. E6.5. Note that the maximum velocit occurs somewhat before the halfwa point in progressing from the inner clinder to the outer clinder. Volumetric flow rate. The final quantit of interest is the volumetric flow rate Q. Observing first that the flow rate through an annulus of internal radius r and external radius r + dr is dq = v z 2πr dr, integration ields: Q Q = dq = v z 2πr dr. (E6.5.9) 0 r 1 Since r ln r is involved in the expression for v z, the following indefinite integral is needed: r ln rdr= r2 r2 ln r 2 4, (E6.5.10) giving the final result: Q = π(r2 2 r 2 1) 8µ r2 ( p )[ ] r r 2 1 r2 2 r1 2. (E6.5.11) z ln(r 2 /r 1 )

22 6.5 Solution of the Equations of Motion in Clindrical Coordinates 293 Since Q, µ, r 1, and r 2 are constant throughout the die, p/ z is also constant, thus verifing the hpothesis previousl made. Observe that in the limiting case of r 1 0, Eqn. (E6.5.11) simplifies to the Hagen-Poiseuille law, alread stated in Eqn. (3.12). This problem ma also be solved b performing a momentum balance on a shell that consists of an annulus of internal radius r, external radius r + dr, and length dz. Example 6.6 Spinning a Polmeric Fiber A Newtonian polmeric liquid of viscosit µ is being spun (drawn into a fiber or filament of small diameter before solidifing b pulling it through a chemical setting bath) in the apparatus shown in Fig. E6.6. The liquid volumetric flow rate is Q, and the filament diameters at z = 0 and z = L are D 0 and D L, respectivel. To a first approximation, the effects of gravit, inertia, and surface tension are negligible. Derive an expression for the tensile force F needed to pull the filament downwards. Assume that the axial velocit profile is flat at an vertical location, so that v z depends onl on z, which is here most convenientl taken as positive in the downwards direction. Also derive an expression for the downwards velocit v z as a function of z. The inset of Fig. E6.6 shows further details of the notation concerning the filament. D 0 z = 0 r Suppl tank Q F L z Circular filament v z A F z = L D L Rollers Inset Setting bath Fig. E6.6 Spinning a polmer filament, whose diameter in relation to its length is exaggerated in the diagram.

23 294 Chapter 6 Solution of Viscous-Flow Problems Solution It is first necessar to determine the radial velocit and hence the pressure inside the filament. From continuit: 1 (rv r ) + v z r r z =0. (E6.6.1) Since v z depends onl on z, its axial derivative is a function of z onl, or dv z /dz = f(z), so that Eqn. (E6.6.1) ma be rearranged and integrated at constant z to give: (rv r ) r = rf(z), rv r = r2 f(z) 2 + g(z). (E6.6.2) But to avoid an infinite value of v r at the centerline, g(z) must be zero, giving: v r = rf(z), 2 v r r = f(z) 2. (E6.6.3) To proceed with reasonable expedienc, it is necessar to make some simplification. After accounting for a primar effect (the difference between the pressure in the filament and the surrounding atmosphere), we assume that a secondar effect (variation of pressure across the filament) is negligible; that is, the pressure does not depend on the radial location. Noting that the external (gauge) pressure is zero everwhere, and appling the first part of Eqn. (5.64) at the free surface: σ rr = p +2µ v r r = p µf(z) =0, or p = µdv z dz. The axial stress is therefore: (E6.6.4) σ zz = p +2µ dv z dz =3µdv z dz. (E6.6.5) It is interesting to note that the same result can be obtained with an alternative assumption. 2 The axial tension in the fiber equals the product of the cross-sectional area and the local axial stress: F = Aσ zz =3µA dv z dz. (E6.6.6) Since the effect of gravit is stated to be insignificant, F is a constant, regardless of the vertical location. At an location, the volumetric flow rate equals the product of the crosssectional area and the axial velocit: Q = Av z. (E6.6.7) 2 See page 235 of S. Middleman s Fundamentals of Polmer Processing, McGraw-Hill, New York, There, the author assumes σ rr = σ θθ = 0, followed b the identit: p = (σ zz + σ rr + σ θθ )/3.

24 6.6 Solution of the Equations of Motion in Spherical Coordinates 295 A differential equation for the velocit is next obtained b dividing one of the last two equations b the other, and rearranging: 1 v z dv z dz = F 3µQ. (E6.6.8) Integration, noting that the inlet velocit at z =0isv z0 = Q/(πD 2 0/4), gives: vz v z0 dv z = F z dz, where v z0 = 4Q, (E6.6.9) v z 3µQ 0 πd0 2 so that the axial velocit obes: v z = v z0 e Fz/3µQ. (E6.6.10) The tension is obtained b appling Eqns. (E6.6.7) and (E6.6.10) just before the filament is taken up b the rollers: Rearrangement ields: v zl = 4Q πd 2 L F = 3µQ L = v z0 e FL/3µQ. (E6.6.11) ln v zl v z0, (E6.6.12) which predicts a force that increases with higher viscosities, flow rates, and drawdown ratios (v zl /v z0 ), and that decreases with longer filaments. Elimination of F from Eqns. (E6.6.10) and (E6.6.12) gives an expression for the velocit that depends onl on the variables specified originall: ( ) z/l vzl v z = v z0 = v z0 v z0 a result that is independent of the viscosit. ( D0 D L ) 2z/L, (E6.6.13) 6.6 Solution of the Equations of Motion in Spherical Coordinates Most of the introductor viscous-flow problems will lend themselves to solution in either rectangular or clindrical coordinates. Occasionall, as in Example 6.7, a problem will arise in which spherical coordinates should be used. It is a fairl advanced problem! Tr first to appreciate the broad steps involved, and then peruse the fine detail at a second reading.

25 296 Chapter 6 Solution of Viscous-Flow Problems Example 6.7 Analsis of a Cone-and-Plate Rheometer The problem concerns the analsis of a cone-and-plate rheometer, an instrument developed and perfected in the 1950s and 1960s b Prof. Karl Weissenberg, for measuring the viscosit of liquids, and also known as the Weissenberg rheogoniometer. 3 A cross section of the essential features is shown in Fig. E6.7, in which the liquid sample is held b surface tension in the narrow opening between a rotating lower circular plate, of radius R, and an upper cone, making an angle of β with the vertical axis. The plate is rotated steadil in the φ direction with an angular velocit ω, causing the liquid in the gap to move in concentric circles with a velocit v φ. (In practice, the tip of the cone is slightl truncated, to avoid friction with the plate.) Observe that the flow is of the Couette tpe. Rigid clamp The torque T needed to hold the cone stationar is measured Upper shaft (torsion bar) R Transducer q = b H Liquid a The plate is rotated with a stead angular velocit w Cone q Axis of rotation Circular plate f direction r q = p /2 Fig. E6.7 Geometr for a Weissenberg rheogoniometer. (The angle between the cone and plate is exaggerated.) The top of the upper shaft which acts like a torsion bar is clamped rigidl. However, viscous friction will twist the cone and the lower portions of the upper shaft ver slightl; the amount of motion can be detected b a light arm at the extremit of which is a transducer, consisting of a small piece of steel, attached to the arm, and surrounded b a coil of wire; b monitoring the inductance of the coil, 3 Professor Weissenberg once related to the author that he (Prof. W.) was attending an instrument trade show in London. There, the rheogoniometer was being demonstrated b a oung salesman who was unaware of Prof. W s identit. Upon inquiring how the instrument worked, the salesman replied: I m sorr, sir, but it s quite complicated, and I don t think ou will be able to understand it.

26 6.6 Solution of the Equations of Motion in Spherical Coordinates 297 the small angle of twist can be obtained; a knowledge of the elastic properties of the shaft then enables the restraining torque T to be obtained. From the analsis given below, it is then possible to deduce the viscosit of the sample. The instrument is so sensitive that if no liquid is present, it is capable of determining the viscosit of the air in the gap! The problem is best solved using spherical coordinates, because the surfaces of the cone and plate are then described b constant values of the angle θ, namel β and π/2, respectivel. Assumptions and the continuit equation. The following realistic assumptions are made: 1. There is onl one nonzero velocit component, namel that in the φ direction, v φ. Thus, v r = v θ =0. 2. Gravit acts verticall downwards, so that g φ =0. 3. We do not need to know how the pressure varies in the liquid. Therefore, the r and θ momentum balances, which would suppl this information, are not required. The analsis starts once more from the continuit equation, (5.50): ρ t + 1 (ρr 2 v r ) + 1 r 2 r r sin θ (ρv θ sin θ) θ + 1 r sin θ (ρv φ ) φ 1 2 v φ r 2 sin 2 θ φ v φ 2 r 2 sin 2 θ + 2 v r r 2 sin θ φ + 2 cos θ r 2 sin 2 θ =0, (5.50) which, for constant densit and v r = v θ = 0 reduces to: v φ φ =0, (E6.7.1) verifing that v φ is independent of the angular location φ, so we are correct in examining just one representative cross section, as shown in Fig. E6.7. Momentum balances. There are again three momentum balances, one for each of the r, θ, and φ directions. From the third assumption above, the first two such balances are of no significant interest, leaving, from Eqn. (5.77), just that in the φ direction: ( vφ ρ t + v v φ r r + v θ v φ r θ + v φ v φ r sin θ φ + v φv r + v ) θv φ cot θ r r = 1 [ ( p 1 r sin θ φ + µ r 2 v ) φ + 1 ( sin θ v ) φ (E6.7.2) r 2 r r r 2 sin θ θ θ ] + + ρg φ. With v r = v θ = 0 (assumption 1), v φ / φ = 0 [from Eqn. (E6.7.1)], and g φ = 0 (assumption 2), the momentum balance simplifies to: ( r 2 v ) φ + 1 ( sin θ v ) φ v φ r r sin θ θ θ sin 2 θ =0. (E6.7.3) v θ φ

27 298 Chapter 6 Solution of Viscous-Flow Problems Determination of the velocit profile. First, seek an expression for the velocit in the φ direction, which is expected to be proportional to both the distance r from the origin and the angular velocit ω of the lower plate. However, its variation with the coordinate θ is something that has to be discovered. Therefore, postulate a solution of the form: v φ = rωf(θ), (E6.7.4) in which the function f(θ) is to be determined. Substitution of v φ from Eqn. (E6.7.4) into Eqn. (E6.7.3), and using f and f to denote the first and second total derivatives of f with respect to θ, gives: f + f cot θ + f(1 cot 2 θ) 1 d sin 2 θ dθ (f sin 2 θ f sin θ cos θ) = 1 [ d sin 3 θ d ( )] f =0. (E6.7.5) sin 2 θ dθ dθ sin θ The reader is encouraged, as alwas, to check the missing algebraic and trigonometric steps, although the are rather trick here! 4 B multipling Eqn. (E6.7.5) through b sin 2 θ, it follows after integration that the quantit in brackets is a constant, here represented as 2c 1, the reason for the 2 being that it will cancel with a similar factor later on: sin 3 θ d ( ) f = 2c 1. (E6.7.6) dθ sin θ Separation of variables and indefinite integration (without specified limits) ields: ( ) f d = f sin θ sin θ = 2c dθ 1 sin 3 θ. (E6.7.7) To proceed further, we need the following two standard indefinite integrals and one trigonometric identit: 5 dθ sin 3 θ = 1 cot θ 2 sin θ + 1 dθ 2 sin θ, (E6.7.8) ( dθ sin θ =ln tan θ ) = 12 ( 2 ln tan 2 θ ), (E6.7.9) 2 tan 2 θ 2 = 1 cos θ 1 + cos θ. (E6.7.10) 4 Although our approach is significantl different from that given on page 98 et seq. of Bird, R.B., Stewart, W.E., and Lightfoot, E.N., Transport Phenomena, Wile, New York (1960), we are indebted to these authors for the helpful hint the gave in establishing the equivalenc expressed in our Eqn. (E6.7.5). 5 From pages 87 (integrals) and 72 (trigonometric identit) of Perr, J.H., ed., Chemical Engineers Handbook, 3rd ed., McGraw-Hill, New York (1950).

28 6.6 Solution of the Equations of Motion in Spherical Coordinates 299 Armed with these, Eqn. (E6.7.7) leads to the following expression for f: f = c 1 [ cot θ in which c 2 is a second constant of integration. ( ln 1 + cos θ ) ] sin θ + c 2, (E6.7.11) 1 cos θ Implementation of the boundar conditions. The constants c 1 and c 2 are found b imposing the two boundar conditions: 1. At the lower plate, where θ = π/2 and the expression in parentheses in Eqn. (E6.7.11) is zero, so that f = c 2, the velocit is simpl the radius times the angular velocit: v φ rωf = rωc 2 = rω, or c 2 =1. (E6.7.12) 2. At the surface of the cone, where θ = β, the velocit v φ = rωf is zero. Hence f = 0, and Eqn. (E6.7.11) leads to: c 1 = c 2 g(β) = g(β), where 1 g(β) = cot β + 1 ( ln 1 + cos β ) sin β. (E6.7.13) 2 1 cos β Substitution of these expressions for c 1 and c 2 into Eqn. (E6.7.11), and noting that v φ = rωf, gives the final (!) expression for the velocit: cot θ + 1 v φ = rω 1 2 cot β ( ln 1 + cos θ ) sin θ 1 cos θ ( ln 1 + cos β ). sin β 1 cos β (E6.7.14) As a partial check on the result, note that Eqn. (E6.7.14) reduces to v φ = rω when θ = π/2 and to v φ = 0 when θ = β. Shear stress and torque. Recall that the primar goal of this investigation is to determine the torque T needed to hold the cone stationar. The relevant shear stress exerted b the liquid on the surface of the cone is τ θφ that exerted on the under surface of the cone (of constant first subscript, θ = β) in the positive φ direction (refer again to Fig for the sign convention and notation for stresses). Since this direction is the same as that of the rotation of the lower plate, we expect that τ θφ will prove to be positive, thus indicating that the liquid is tring to turn the cone in the same direction in which the lower plate is rotated. From the second of Eqn. (5.65), the relation for this shear stress is: [ sin θ τ θφ = µ r ( vφ ) + 1 θ sin θ r sin θ ] v θ. (E6.7.15) φ

29 300 Chapter 6 Solution of Viscous-Flow Problems Since v θ = 0, and recalling Eqns. (E6.7.4), (E6.7.6), and (E6.7.13), the shear stress on the cone becomes: [ sin θ (τ θφ ) θ=β = µ r ( )] rωf = θ sin θ θ=β ( ) 2c1 ωµ = 2ωµg(β) sin 2 θ θ=β sin 2 β. (E6.7.16) One importance of this result is that it is independent of r, giving a constant stress and strain throughout the liquid, a significant simplification when deciphering the experimental results for non-newtonian fluids (see Chapter 11). In effect, the increased velocit differences between the plate and cone at the greater values of r are offset in exact proportion b the larger distances separating them. The torque exerted b the liquid on the cone (in the positive φ direction) is obtained as follows. The surface area of the cone between radii r and r + dr is 2πr sin βdr, and is located at a lever arm of r sin β from the axis of smmetr. Multiplication b the shear stress and integration gives: T = R/ sin β 0 (2πr sin βdr) } {{ } Area r sin β } {{ } Lever arm (τ θφ ) θ=β, (E6.7.17) } {{ } Stress Substitution of (τ θφ ) θ=β from Eqn. (E6.7.16) and integration gives the torque as: T = 4πωµg(β)R3. (E6.7.18) 3 sin 3 β The torque for holding the cone stationar has the same value, but is, of course, in the negative φ direction. Since R and g(β) can be determined from the radius and the angle β of the cone in conjunction with Eqn. (E6.7.13), the viscosit µ of the liquid can finall be determined. Problems for Chapter 6 Unless otherwise stated, all flows are stead state, for a Newtonian fluid with constant densit and viscosit. 1. Stretching of a liquid film M. In broad terms, explain the meanings of the following two equations, paing attention to an sign convention: σ xx = p +2µ v x x, σ = p +2µ v.

30 Problems for Chapter Bar Viscous liquid film Bar F F z x Fig. P6.1 L Stretching of liquid film between two bars. Fig. P6.1 shows a film of a viscous liquid held between two bars spaced a distance L apart. If the film thickness is uniform, and the total volume of liquid is V, show that the force necessar to separate the bars with a relative velocit dl/dt is: F = 4µV L 2 2. Wire coating M. Fig. P6.2 shows a rodlike wire of radius r 1 that is being pulled steadil with velocit V through a horizontal die of length L and internal radius r 2. The wire and the die are coaxial, and the space between them is filled with a liquid of viscosit µ. The pressure at both ends of the die is atmospheric. The wire is coated with the liquid as it leaves the die, and the thickness of the coating eventuall settles down to a uniform value, δ. L dl dt. Film thickness d Wire r 2 r 1 Velocit V Fig. P6.2 Die Coating of wire drawn through a die. Neglecting end effects, use the equations of motion in clindrical coordinates to derive expressions for: (a) The velocit profile within the annular space. Assume that there is onl one nonzero velocit component, v z, and that this depends onl on radial position. (b) The total volumetric flow rate Q through the annulus. (c) The limiting value for Q if r 1 approaches zero. (d) The final thickness, δ, of the coating on the wire. (e) The force F needed to pull the wire.

31 302 Chapter 6 Solution of Viscous-Flow Problems 3. Off-center annular flow D (C). A liquid flows under a pressure gradient p/ z through the narrow annular space of a die, a cross section of which is shown in Fig. P6.3(a). The coordinate z is in the axial direction, normal to the plane of the diagram. The die consists of a solid inner clinder with center P and radius b inside a hollow outer clinder with center O and radius a. The points O and P were intended to coincide, but due to an imperfection of assembl are separated b a small distance δ. a D b O q d P Inner clinder Q Outer clinder P q b D dq b dq (a) Fig. P6.3 Off-center clinder inside a die (gap width exaggerated): (a) complete cross section; (b) effect of incrementing θ. B a simple geometrical argument based on the triangle OPQ, show that the gap width between the two clinders is given approximatel b: =. a b δ cos θ, where the angle θ is defined in the diagram. Now consider the radius arm b swung through an angle dθ, so that it traces an arc of length bdθ. The flow rate dq through the shaded element in Fig. P6.3(b) is approximatel that between parallel plates of width bdθ and separation. Hence prove that the flow rate through the die is given approximatel b: Q = πbc(2α 3 +3αδ 2 ), in which: c = 1 ( p ), and α = a b. 12µ z Assume from Eqn. (E6.1.26) that the flow rate per unit width between two flat plates separated b a distance h is: ( h 3 p ). 12µ z What is the ratio of the flow rate if the two clinders are touching at one point to the flow rate if the are concentric? (b)

32 Problems for Chapter Compression molding M. Fig. P6.4 shows the (a) beginning, (b) intermediate, and (c) final stages in the compression molding of a material that behaves as a liquid of high viscosit µ, from an initial clinder of height H 0 and radius R 0 to a final disk of height H 1 and radius R 1. V A H 0 H H 1 B R 0 R R 1 (a) (b) (c) Fig. P6.4 Compression molding between two disks. In the molding operation, the upper disk A is squeezed with a uniform velocit V towards the stationar lower disk B. Ignoring small variations of pressure in the z direction, prove that the total compressive force F that must be exerted downwards on the upper disk is: 3πµV R4 F =. 2H 3 Assume that the liquid flow is radiall outwards everwhere, with a parabolic velocit profile. Also assume from Eqn. (E6.1.26) that per unit width of a channel of depth H, the volumetric flow rate is: Q = H ( 3 p ). 12µ r Give expressions for H and R as functions of time t, and draw a sketch that shows how F varies with time. 5. Film draining M. Fig. P6.5 shows an idealized view of a liquid film of viscosit µ that is draining under gravit down the side of a flat vertical wall. Such a situation would be approximated b the film left on the wall of a tank that was suddenl drained through a large hole in its base. What are the justifications for assuming that the velocit profile at an distance x below the top of the wall is given b: v x = ρg (2h ), 2µ where h = h(x) is the local film thickness? Derive an expression for the corresponding downwards mass flow rate m per unit wall width (normal to the plane of the diagram).

33 304 Chapter 6 Solution of Viscous-Flow Problems Perform a transient mass balance on a differential element of the film and prove that h varies with time and position according to: x h t = 1 m ρ x. h v x Wall Liquid film Fig. P6.5 Liquid draining from a vertical wall. Now substitute our expression for m, to obtain a partial differential equation for h. Tr a solution of the form: h = ct p x q, and determine the unknowns c, p, and q. Discuss the limitations of our solution. Note that a similar situation occurs when a substrate is suddenl lifted from a bath of coating fluid. 6. Sheet spinning M. A Newtonian polmeric liquid of viscosit µ is being spun (drawn into a sheet of small thickness before solidifing b pulling it through a chemical setting bath) in the apparatus shown in Fig. P6.6. The liquid volumetric flow rate is Q, and the sheet thicknesses at z = 0 and z = L are and δ, respectivel. The effects of gravit, inertia, and surface tension are negligible. Derive an expression for the tensile force needed to pull the filament downwards. Hint: start b assuming that the verticall downwards velocit v z depends onl on z and that the lateral velocit v is zero. Also derive an expression for the downwards velocit v z as a function of z.

34 Problems for Chapter z = 0 z = L D v z z d x W Polmer sheet Roller Suppl tank Fig. P6.6 Spinning a polmer sheet. 7. Details of pipe flow M. A fluid of densit ρ and viscosit µ flows from left to right through the horizontal pipe of radius a and length L shown in Fig. P6.7. The pressures at the centers of the inlet and exit are p 1 and p 2, respectivel. You ma assume that the onl nonzero velocit component is v z, and that this is not a function of the angular coordinate, θ. Wall Inlet r a p 1 p 2 z Exit Axis of smmetr Fig. P6.7 L Flow of a liquid in a horizontal pipe. Stating an further necessar assumptions, derive expressions for the following, in terms of an or all of a, L, p 1, p 2, ρ, µ, and the coordinates r, z, and θ: (a) The velocit profile, v z = v z (r). (b) The total volumetric flow rate Q through the pipe.

35 306 Chapter 6 Solution of Viscous-Flow Problems (c) The pressure p at an point (r, θ, z). (d) The shear stress, τ rz. 8. Natural convection M. Fig. P6.8 shows two infinite parallel vertical walls that are separated b a distance 2d. A fluid of viscosit µ and volume coefficient of expansion β fills the intervening space. The two walls are maintained at uniform temperatures T 1 (cold) and T 2 (hot), and ou ma assume (to be proved in a heattransfer course) that there is a linear variation of temperature in the x direction. That is: T = T + x ( ) T2 T 1, where T = T 1 + T 2. d 2 2 The densit is not constant, but varies according to: ρ = ρ [ 1 β ( T T )], where ρ is the densit at the mean temperature T, which occurs at x =0. v T 1 Cold wall T2 Hot wall Fig. P6.8 x = - d x x = d Natural convection between vertical walls. If the resulting natural-convection flow is stead, use the equations of motion to derive an expression for the velocit profile v = v (x) between the plates. Your expression for v should be in terms of an or all of x, d, T 1, T 2, ρ, µ, β, and g. Hints: in the momentum balance, ou should find ourself facing the following combination: p + ρg, in which g = g. These two terms are almost in balance, but not quite, leading to a small but important buoanc effect that drives the natural convection.

36 Problems for Chapter The variation of pressure in the direction ma be taken as the normal hdrostatic variation: p = gρ. We then have: p + ρg = gρ + ρ [ 1 β ( T T )] ( g) =βg ( T T ), and this will be found to be a vital contribution to the momentum balance. 9. Square duct velocit profile M. A certain flow in rectangular Cartesian coordinates has onl one nonzero velocit component, v z, and this does not var with z. If there is no bod force, write down the Navier-Stokes equation for the z momentum balance. = a x x = - a = - a x = a Fig. P6.9 Square cross section of a duct. One-dimensional, full developed stead flow occurs under a pressure gradient p/ z in the z direction, parallel to the axis of a square duct of side 2a, whose cross section is shown in Fig. P6.9. The following equation has been proposed for the velocit profile: v z = 1 ( p ) [ ( x ) 2 ][ ( 2 ] a µ z a a) Without attempting to integrate the momentum balance, investigate the possible merits of this proposed solution for v z. Explain whether or not it is correct. = a x = - a x = - a x = a Fig. P6.10 Square cross section of a duct.

37 308 Chapter 6 Solution of Viscous-Flow Problems 10. Poisson s equation for a square duct E. A polmeric fluid of uniform viscosit µ is to be extruded after pumping it through a long duct whose cross section is a square of side 2a, shown in Fig. P6.10. The flow is parallel everwhere to the axis of the duct, which is in the z direction, normal to the plane of the diagram. If p/ z, µ, and a are specified, show that the problem of obtaining the axial velocit distribution v z = v z (x, ), amounts to solving Poisson s equation of the form 2 φ = f(x, ), where f is specified and φ is the unknown. Also note that Poisson s equation can be solved numericall b the Matlab PDE (partial differential equation) Toolbox, as outlined in Chapter 12. Wall Wall A A Centerline B B x (a) (b) Wall A Wall A Centerline B B x Fig. P6.11 (c) (d) Proposed velocit profiles for immiscible liquids. 11. Permissible velocit profiles E. Consider the shear stress τ x ; wh must it be continuous in the direction, for example and not undergo a sudden stepchange in its value? Two immiscible Newtonian liquids A and B are in stead laminar flow between two parallel plates. Profile A meets the centerline normall in (b), but at an angle in (c); the maximum velocit in (d) does not coincide with the centerline. Which if an of the velocit profiles shown in Fig. P6.11 are impossible? Explain our answers carefull. 12. Creeping flow past a sphere D. Figure P6.12 shows the stead, creeping (ver slow) flow of a fluid of viscosit µ past a sphere of radius a. Far awa from the sphere, the pressure is p and the undisturbed fluid velocit is U in the positive z direction. The following velocit components and pressure have been

38 Problems for Chapter proposed in spherical coordinates: p = p 3µUa cos θ, ( 2r 2 v r = U 1 3a ) 2r + a3 cos θ, 2r ( 3 v θ = U 1 3a ) 4r a3 sin θ, 4r 3 v φ =0. Assuming the velocities are sufficientl small so that terms such as v r ( v r / r) can be neglected, and that gravit is unimportant, prove that these equations do indeed satisf the following conditions, and therefore are the solution to the problem: (a) The continuit equation. (b) The r and θ momentum balances. (c) A pressure of p and a z velocit of U far awa from the sphere. (d) Zero velocit components on the surface of the sphere. U, p r O a q z Fig. P6.12 Viscous flow past a sphere. Also derive an expression for the net force exerted in the z direction b the fluid on the sphere, and compare it with that given b Stokes law in Eqn. (4.11). Note that the problem is one in spherical coordinates, in which the z axis has no formal place, except to serve as a reference direction from which the angle θ is measured. There is also smmetr about this axis, such that an derivatives in the φ direction are zero. Note: the actual derivation of these velocities, starting from the equations of motion, is fairl difficult!

39 310 Chapter 6 Solution of Viscous-Flow Problems 13. Torque in a Couette viscometer M. Fig. P6.13 shows the horizontal cross section of a concentric clinder or Couette viscometer, which is an apparatus for determining the viscosit µ of the fluid that is placed between the two vertical clinders. The inner and outer clinders have radii of r 1 and r 2, respectivel. If the inner clinder is rotated with a stead angular velocit ω, and the outer clinder is stationar, derive an expression for v θ (the θ velocit component) as a function of radial location r. Inner clinder w r O q Outer clinder r 1 r 2 Fig. P6.13 Section of a Couette viscometer. If, further, the torque required to rotate the inner clinder is found to be T per unit length of the clinder, derive an expression whereb the unknown viscosit µ can be determined, in terms of T, ω, r 1, and r 2. Hint: ou will need to consider one of the shear stresses given in Table 5.8. Flow rate Q Wall Liquid film x d Fig. P6.14 Wetted-wall column. 14. Wetted-wall column M. Fig. P6.14 shows a wetted-wall column, in which a thin film of a reacting liquid of viscosit µ flows steadil down a plane

40 Problems for Chapter wall, possibl for a gas-absorption stud. The volumetric flow rate of liquid is specified as Q per unit width of the wall (normal to the plane of the diagram). Assume that there is onl one nonzero velocit component, v x, and that this does not var in the x direction, and that the gas exerts negligible shear stress on the liquid film. Starting from the equations of motion, derive an expression for the profile of the velocit v x (as a function of ρ, µ, g,, and δ), and also for the film thickness, δ (as a function of ρ, µ, g, and Q). 15. Simplified view of a Weissenberg rheogoniometer M. Consider the Weissenberg rheogoniometer with a ver shallow cone; thus, referring to Fig. E6.7, β = π/2 α, where α is a small angle. (a) Without going through the complicated analsis presented in Example 6.7, outline our reasons for supposing that the shear stress at an location on the cone is: (τ θφ ) θ=β = ωµ α. (b) Hence, prove that the torque required to hold the cone stationar (or to rotate the lower plate) is: T = 2 πωµr 4 3 H. (c) B substituting β = π/2 α into Eqn. (E6.7.13) and expanding the various functions in power series (onl a ver few terms are needed), prove that g(β) = 1/(2α), and that Eqn. (E6.7.18) again leads to the expression just obtained for the torque in part (b) above. Circular disk Rigid clamp R H Liquid z r The lower disk is rotated with a stead angular velocit w Circular disk q direction Fig. P6.16 Axis of rotation Cross section of parallel-disk rheometer.

41 312 Chapter 6 Solution of Viscous-Flow Problems 16. Parallel-disk rheometer M. Fig. P6.16 shows the diametral cross section of a viscometer, which consists of two opposed circular horizontal disks, each of radius R, spaced b a vertical distance H; the intervening gap is filled b a liquid of constant viscosit µ and constant densit. The upper disk is stationar, and the lower disk is rotated at a stead angular velocit ω in the θ direction. There is onl one nonzero velocit component, v θ, so the liquid everwhere moves in circles. Simplif the general continuit equation in clindrical coordinates, and hence deduce those coordinates (r?, θ?, z?) on which v θ ma depend. Now consider the θ-momentum equation, and simplif it b eliminating all zero terms. Explain briefl: (a) wh ou would expect p/ θ to be zero, and (b) wh ou cannot neglect the term 2 v θ / z 2. Also explain briefl the logic of supposing that the velocit in the θ direction is of the form v θ = rωf(z), where the function f(z) is et to be determined. Now substitute this into the simplified θ-momentum balance and determine f(z), using the boundar conditions that v θ is zero on the upper disk and rω on the lower disk. Wh would ou designate the shear stress exerted b the liquid on the lower disk as τ zθ? Evaluate this stress as a function of radius. 17. Screw extruder optimum angle M. Note that the flow rate through the die of Example 6.5, given in Eqn. (E6.5.10), can be expressed as: Q = c(p 2 p 3 ) µd, in which c is a factor that accounts for the geometr. Suppose that this die is now connected to the exit of the extruder studied in Example 6.4, and that p 1 = p 3 = 0, both pressures being atmospheric. Derive an expression for the optimum flight angle θ opt that will maximize the flow rate Q through the extruder and die. Give our answer in terms of an or all of the constants c, D, h, L 0, r, W, µ, and ω. Under what conditions would the pressure at the exit of the extruder have its largest possible value p 2max? Derive an expression for p 2max. 18. Annular flow in a die E. Referring to Example 6.5, concerning annular flow in a die, answer the following questions, giving our explanation in both cases: (a) What form does the velocit profile, v z = v z (r), assume as the radius r 1 of the inner clinder becomes vanishingl small? (b) Does the maximum velocit occur halfwa between the inner and outer clinders, or at some other location? 19. Rotating rod in a fluid M. Fig. P6.19(a) shows a horizontal cross section of a long vertical clinder of radius a that is rotated steadil counterclockwise with an angular velocit ω in a ver large volume of liquid of viscosit µ. The liquid extends effectivel to infinit, where it ma be considered at rest. The axis of the clinder coincides with the z axis.

42 Problems for Chapter A r a w q w B (a) Fig. P6.19 Rotating clinder in (a) a single liquid, and (b) two immiscible liquids. (a) What tpe of flow is involved? What coordinate sstem is appropriate? (b) Write down the differential equation of mass and that one of the three general momentum balances that is most applicable to the determination of the velocit v θ. (c) Clearl stating our assumptions, simplif the situation so that ou obtain an ordinar differential equation with v θ as the dependent variable and r as the independent variable. (d) Integrate this differential equation, and introduce an boundar condition(s), and prove that v θ = ωa 2 /r. (e) Derive an expression for the shear stress τ rθ at the surface of the clinder. Carefull explain the plus or minus sign in this expression. (f) Derive an expression that gives the torque T needed to rotate the clinder, per unit length of the clinder. (g) Derive an expression for the vorticit component ( v) z. Comment on our result. (h) Fig. P6.19(b) shows the initial condition of a mixing experiment in which the clinder is in the middle of two immiscible liquids, A and B, ofidentical densities and viscosities. After the clinder has made one complete rotation, draw a diagram that shows a representative location of the interface between A and B. 20. Two-phase immiscible flow M. Fig. P6.20 shows an apparatus for measuring the pressure drop of two immiscible liquids as the flow horizontall between two parallel plates that extend indefinitel normal to the plane of the diagram. The liquids, A and B, have viscosities µ A and µ B, densities ρ A and ρ B, and volumetric flow rates Q A and Q B (per unit depth normal to the plane of the figure), respectivel. Gravit ma be considered unimportant, so that the pressure is essentiall onl a function of the horizontal distance, x. (b)

43 314 Chapter 6 Solution of Viscous-Flow Problems x = 0 B x = L = H = d A x Fig. P6.20 Two-phase flow between parallel plates. = 0 (a) What tpe of flow is involved? (b) Considering laer A, start from the differential equations of mass and momentum, and, clearl stating our assumptions, simplif the situation so that ou obtain a differential equation that relates the horizontal velocit v xa to the vertical distance. (c) Integrate this differential equation so that ou obtain v xa in terms of and an or all of d, dp/dx, µ A, ρ A, and (assuming the pressure gradient is uniform) two arbitrar constants of integration, sa, c 1A and c 2A. Assume that a similar relationship holds for v xb. (d) Clearl state the four boundar and interfacial conditions, and hence derive expressions for the four constants, thus giving the velocit profiles in the two laers. (e) Sketch the velocit profiles and the shear-stress distribution for τ x between the upper and lower plates. (f) Until now, we have assumed that the interface level = d is known. In realit, however, it will depend on the relative flow rates Q A and Q B. Show clearl how this dependenc could be obtained, but do not actuall carr the calculations through to completion. Drops Film Axis of rotation z r w Impeller Fig. P6.21 Rotating-impeller humidifier. 21. Room humidifier M. Fig. P6.21 shows a room humidifier, in which a circular impeller rotates about its axis with angular velocit ω. A conical exten-

44 Problems for Chapter sion dips into a water-bath, sucking up the liquid (of densit ρ and viscosit µ), which then spreads out over the impeller as a thin laminar film that rotates everwhere with an angular velocit ω, eventuall breaking into drops after leaving the peripher. (a) Assuming incompressible stead flow, with smmetr about the vertical (z) axis, and a relativel small value of v z, what can ou sa from the continuit equation about the term: (rv r )? r (b) If the pressure in the film is everwhere atmospheric, the onl significant inertial term is vθ/r, 2 and information from (a) can be used to neglect one particular term, to what two terms does the r momentum balance simplif? (c) Hence prove that the velocit v r in the radial direction is a half-parabola in the z direction. (d) Derive an expression for the total volumetric flow rate Q, and hence prove that the film thickness at an radial location is given b: ( ) 1/3 3Qν δ =, 2πr 2 ω 2 where ν is the kinematic viscosit. Gap width is exaggerated here Q r c a Fig. P6.22 v c Transport of inner clinder. 22. Transport of inner clinder M (C). As shown in Fig. P6.22, a long solid clinder of radius r c and length L is being transported b a viscous liquid of the same densit down a pipe of radius a, which is much smaller than L. The annular gap, of extent a r c, is much smaller than a. Assume: (a) the clinder remains concentric within the pipe, (b) the flow in the annular gap is laminar, (c) the shear stress is essentiall constant across the gap, and (d) entr and exit effects can be neglected. Prove that the velocit of the clinder is given fairl accuratel b: v c = 2Q π(a 2 + r 2 c), where Q is the volumetric flow rate of the liquid upstream and downstream of the clinder. Hint: concentrate first on understanding the phsical situation. Don t rush headlong into a length analsis with the Navier-Stokes equations!

45 316 Chapter 6 Solution of Viscous-Flow Problems r q Flow a a vz z Fig. P6.23 Flow between inclined planes. 23. Flow between inclined planes M (C). A viscous liquid flows between two infinite planes inclined at an angle 2α to each other. Prove that the liquid velocit, which is everwhere parallel to the line of intersection of the planes, is given b: ( )( v z = r2 cos 2θ 4µ cos 2α 1 p ), z where z, r, θ are clindrical coordinates. The z-axis is the line of intersection of the planes and the r-axis (θ = 0) bisects the angle between the planes. Assume laminar flow, with: [ p 1 z = µ 2 v z = µ r r ( r v ) z + 1 ] 2 v z, r r 2 θ 2 and start b proposing a solution of the form v z = r n g(θ)( p/ z)/µ, where the exponent n and function g(θ) are to be determined. 24. Immiscible flow inside a tube D (C). A film of liquid of viscosit µ 1 flows down the inside wall of a circular tube of radius (λ + ). The central core is occupied b a second immiscible liquid of viscosit µ 2, in which there is no net vertical flow. End effects ma be neglected, and stead-state circulation in the core liquid has been reached. If the flow in both liquids is laminar, so that the velocit profiles are parabolic as shown in Fig. P6.24, prove that: α = 2µ λµ 1 4µ 2 +λµ 1, where is the thickness of the liquid film, and α is the distance from the wall to the point of maximum velocit in the film. Fig. P6.24 suggests notation for solving the problem. To save time, assume parabolic velocit profiles without proof: u = a + bx + cx 2 and v = d + e + f 2.

46 a x u D v m 1 m 2 Problems for Chapter Fig. P6.24 Flow of two immiscible liquids in a pipe. The thickness of the film next to the wall is exaggerated. Discuss what happens when λ/ ; and also when µ 2 /µ 1 0; and when µ 1 /µ Blowing a polethlene bubble D. For an incompressible fluid in clindrical coordinates, write down: (a) The continuit equation, and simplif it. (b) Expressions for the viscous normal stresses σ rr and σ θθ, in terms of pressure, viscosit, and strain rates. P l t O q a s qq P +Dp Fig. P6.25 Cross section of half of a clindrical bubble. A polethlene sheet is made b inflating a clindrical bubble of molten polmer effectivel at constant length, a half cross section of which is shown in Fig. P6.25. The excess pressure inside the bubble is small compared with the external. pressure P, so that p P and σ rr = P. B means of a suitable force balance on the indicated control volume, prove that the circumferential stress is given b σ θθ = P + a p/t. Assume pseudostead state that is, the circumferential stress just balances the excess pressure, neglecting an acceleration effects. s qq

47 318 Chapter 6 Solution of Viscous-Flow Problems Hence, prove that the expansion velocit v r of the bubble (at r = a) is given b: v r = a2 p 4µt, and evaluate it for a bubble of radius 1.0 m and film thickness 1 mm when subjected to an internal gauge pressure of p = 40 N/m 2. The viscosit of polethlene at the appropriate temperature is 10 5 N s/m Surface-tension effect in spinning M. Example 6.6 ignored surfacetension effects, which would increase the pressure in a filament of radius R approximatel b an amount σ/r, where σ is the surface tension. Compare this quantit with the reduction of pressure, µdv z /dz, caused b viscous effects, for a polmer with µ =10 4 P, σ =0.030 kg/s 2, L =1m,R L (exit radius) = m, v z0 =0.02 m/s, and v zl = 2 m/s. Consider conditions both at the beginning and end of the filament. Comment briefl on our findings. 27. Radial pressure variations in spinning M. Example 6.6 assumed that the variation of pressure across the filament was negligible. Investigate the validit of this assumption b starting with the suitabl simplified momentum balance: [ ( ) ] p 1 r = µ (rv r ) + 2 v r. r r r z 2 If v z is the local axial velocit, prove that the corresponding increase of pressure from just inside the free surface (p R ) to the centerline (p 0 ) is: p 0 p R = µv z(ln β) 3 R 2, 4L 3 where β = v zl /v z0. Obtain an expression for the ratio ξ =(p R p 0 )/(µdv z /dz), in which the denominator is the pressure decrease due to viscosit when crossing the interface from the air into the filament, and which was accounted for in Example 6.6. Estimate ξ at the beginning of the filament for the situation in which µ =10 4 P, L =1m,R L (exit radius) = m, v z0 =0.02 m/s, and v zl = 2 m/s. Comment briefl on our findings. 28. Condenser with varing viscosit M. In a condenser, a viscous liquid flows steadil under gravit as a uniform laminar film down a vertical cooled flat plate. Due to conduction, the liquid temperature T varies linearl across the film, from T 0 at the cooled plate to T 1 at the hotter liquid/vapor interface, according to T = T 0 + (T 1 T 0 )/h. The viscosit of the liquid is given approximatel b µ = µ (1 αt ), where µ and α are constants. Prove that the viscosit at an location can be reexpressed as: µ = µ 0 + c, in which c = µ 1 µ 0, h

48 Problems for Chapter where µ 0 and µ 1 are the viscosities at temperatures T 0 and T 1, respectivel. What is the expression for the shear stress τ x for a liquid for stead flow in the x-direction? Derive an expression for the velocit v x as a function of. Make sure that ou do not base our answer on an equations that assume constant viscosit. Sketch the velocit profile for both a small and a large value of α. 29. True/false. Check true or false, as appropriate: (a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearl both in the direction of flow and in the direction normal to the plates. T F (b) For horizontal flow of a liquid in a rectangular duct between parallel plates, the boundar conditions can be taken as zero velocit at one of the plates and either zero velocit at the other plate or zero velocit gradient at the centerline. T F (c) For horizontal flow of a liquid in a rectangular duct between parallel plates, the shear stress varies from zero at the plates to a maximum at the centerline. T F (d) For horizontal flow of a liquid in a rectangular duct between parallel plates, a measurement of the pressure gradient enables the shear-stress distribution to be found. T F (e) In fluid mechanics, when integrating a partial differential equation, ou get one or more constants of integration, whose values can be determined from the boundar condition(s). T F (f) For flows occurring between r = 0 and r = a in clindrical coordinates, the term ln r ma appear in the final expression for one of the velocit components. T F (g) For flows in ducts and pipes, the volumetric flow rate can be obtained b differentiating the velocit profile. T F (h) Natural convection is a situation whose analsis depends on not taking the densit as constant everwhere. T F (i) A ke feature of the Weissenberg rheogoniometer is the fact that a conical upper surface results in a uniform velocit gradient between the cone and the plate, for all values of radial distance. T F

49 320 Chapter 6 Solution of Viscous-Flow Problems (j) If, in three dimensions, the pressure obes the equation p/ = ρg, and both p/ x and p/ z are nonzero, then integration of this equation gives the pressure as p = ρg + c, where c is a constant. T F (k) If two immiscible liquids A and B are flowing in the x direction between two parallel plates, both the velocit v x and the shear stress τ x are continuous at the interface between A and B, where the coordinate is normal to the plates. T F (l) (m) (n) (o) In compression molding of a disk between two plates, the force required to squeeze the plates together decreases as time increases. For flow in a wetted-wall column, the pressure increases from atmospheric pressure at the gas/liquid interface to a maximum at the wall. For one-dimensional flow in a pipe either laminar or turbulent the shear stress τ rz varies linearl from zero at the wall to a maximum at the centerline. In Example 6.1, for flow between two parallel plates, the shear stress τ x is negative in the upper half (where > 0), meaning that phsicall it acts in the opposite direction to that indicated b the convention. T T T T F F F F